# If two functions are equal to their Newton series, is their composition also equal to its Newton series?

Suppose we have two real-valued functions $f(x)$ and $g(x)$, both equal to their Newton series expansion:

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$

$$g(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k g\left (0\right)$$

Is their composition $F(x)=f(g(x))$ also equal to its Newton series expansion (if it converges)?

$$F(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k F\left (0\right)$$

• Please use at least one top-level tag (those with a two-letter prefix). – user9072 Jun 15 '14 at 12:45
• This question is cross-posted from M.SE - please add the link to that question. – user62675 Jun 15 '14 at 13:20
• Really, what's the point of asking all questions of the type "Does the class of functions representable by their Newton series possess natural property P?" in some random order? The answer to this one is "No, as usual". Take any entire $f$ of order less than $1$ that vanishes at every square of an integer and compose with $g(x)=x^2$. – fedja Jun 15 '14 at 18:13
• @fedja interesting, but at what conditions it holds then? – Anixx Jun 15 '14 at 18:40
• Regarding the function fedja asks for: Take $f(z) = \sqrt{z} \sin (2 \pi \sqrt{z})$. This is entire, since it is easy to check that $\sum_{n=0}^{\infty} (-1)^n (2 \pi)^{2n+1} z^{n+1}/(2n+1)!$ is everywhere convergent; we have the easy bound $|\sqrt{z} \sin(2 \pi \sqrt{z})| < \sqrt{|z|} e^{2 \pi \sqrt{|z|}}$; and $f$ clearly vanishes at squares of integers. – David E Speyer Jun 17 '14 at 20:01

Here goes, as promised.

Let $f$ be entire of order less than $1$, so $|f(z)|\le Ce^{|z|^p}$, $p<1$. Write the Newton polynomial $$P(x)=\sum_{k=0}^n\Delta^kf(0) {x \choose k}$$ Note that $g(k)=f(k)-P(k)=0$ for $k=0,1,\dots,n$. On the other hand, we can crudely estimate $|g|$ in a disk of radius $R>2n$ by $Ce^{R^p}+\sum_{k=0}^n(2R)^k\frac 1{k!}|\Delta^k f(0)|$. Now, $\frac 1{k!}|\Delta^k f(0)|\le \max_{[0,R/2]}\frac{|f^{(k)}|}{k!}\le (2/R)^k Ce^{R^p}$ by Cauchy, so we finally get $$|g|\le C 4^n e^{R^p}$$ in the disk of radius $R$ centered at the origin.

Now, for $|x|<n$, each corresponding Blaschke factor $\frac{R(x-k)}{R^2-kx}$ is at most $\frac{3n}R$ in absolute value, so $$|g(x)|\le C\left(\frac{12}{R/n}\right)^n e^{R^p}$$ Choosing $R=n^{1/p}$, we get $|g(x)|\le \left(12en^{-\frac{1-p}p}\right)^n\to 0$ as $n\to\infty$.

• I have accepted this answer, although I would prefer it be posted in the linked question. Can you please tell, did you derive this theorem yourself or had you seen it before? – Anixx Jun 17 '14 at 20:22
• Sorry I would not ask this, should I know that you're a renowned mathematician. Still your proof is explained in too difficult language at least for me to comprehend. – Anixx Jun 17 '14 at 21:17
• @Anixx Neither, as usual. The correct statement is that I've seen enough stuff before to put it together in a few minutes from readily available blocks. As to "renowned mathematician", that is an obvious overstatement. "Someone who got a decent education as a student and still remembers about half of it" is a more accurate description. :-) – fedja Jun 17 '14 at 21:22
• I doubt that is accurate for someone who has theorems in textbooks named after him. – Anixx Jun 17 '14 at 21:23
• So it is good to know that this somehow touches your prevuous research. – Anixx Jun 17 '14 at 21:24

Another counter-example is extractable from Gerald Edgar's answer to this question, where he shows that $\sin (ax)$ is discrete analytic for $a \in (-\pi/3, \pi/3)$. So take $f(x) = \sin ((\pi/4) x)$ (for example) and $g(x) = 9 x$ (for example). Then $f$ and $g$ are discrete analytic but the Newton series for $(f \circ g)(x) = \sin ((9 \pi/4) x)$ will converge to $\sin( (\pi/4) x)$, which equals $\sin ((9 \pi/4) x)$ for integer $x$, but not in general.